(a)
Differentiating with respect to x we have:
\displaystyle \frac x {18} + \frac {2y} {25} \frac {\mathrm dy} {\mathrm dx} = 0
so:
\displaystyle \frac {2y} {25} \frac {\mathrm dy} {\mathrm dx} = -\frac x {18}
giving:
\displaystyle \frac {\mathrm dy} {\mathrm dx} = -\left(\frac {25} {36}\right) \left(\frac x y\right)
So at the point (6 \cos \theta, 5 \sin \theta) we have:
\displaystyle \frac {\mathrm dy} {\mathrm dx} = -\left(\frac {25} {36}\right) \left(\frac {6 \cos \theta} {5 \sin \theta}\right) = -\frac {5 \cos \theta} {6 \sin \theta}
So the gradient of the normal at P is:
\displaystyle \frac {6 \sin \theta} {5 \cos \theta}
So the equation of l is:
\displaystyle y - 5 \sin \theta = \frac {6 \sin \theta} {5 \cos \theta} (x - 6 \cos \theta)
multiplying through 5 \cos \theta we have:
5y \cos \theta - 25 \cos \theta \sin \theta = 6x \sin \theta - 36 \cos \theta
Rearranging we get:
6x \sin \theta - 5y \cos \theta = 11 \sin \theta \cos \theta
(b)
At the point Q we have y = 0, so:
6x \sin \theta = 11 \sin \theta \cos \theta
Since 0 < \theta < \dfrac \pi 2 we have \sin \theta \ne 0 we have:
\displaystyle x = \frac {11 \cos \theta} 6
so:
\displaystyle Q = \left(0, \frac {11 \cos \theta} 6\right)
giving:
\displaystyle |OQ| = \frac {11 \cos \theta} 6
The perpendicular from P to x-axis is the line drawn straight down from P to the x-axis. The foot of this line is its intersection with the x-axis, so R = (6 \cos \theta, 0) meaning that |OR| = 6 \cos \theta.
So:
\displaystyle \frac {|OQ|} {|OR|} = \frac {11 \cos \theta} 6 \times \frac 1 {6 \cos \theta} = \frac {11} {36}
The eccentricity of the ellipse e satisfies:
25 = 36 (1 - e^2)
So:
\displaystyle e^2 = 1 - \frac {25} {36} = \frac {11} {36} = \frac {|OQ|} {|OR|}
as required.
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